$X_n \rightarrow_{a.s.} X$ versus $X_n=X$ wpa 1 Consider a sequence of real-valued random variables $\{X_n\}_{\forall n \in \mathbb{N}}$ and a real-valued random variable $X$
All r.v. are defined on the probability space $
(\Omega, \mathcal{F}, P)$
Could explain what is the relation (equivalent, one implies the other, etc) between 
$$
(1) \hspace{1cm}X_n \rightarrow_{a.s.} X \text{ as $n\rightarrow \infty$}
$$
and 
$$
(2) \hspace{1cm}X_n =X \text{ with probability approaching 1 as $n\rightarrow \infty$}
$$

Some considerations: 
Using the definitions, 
(1) $P(\omega \in \Omega \text{ s.t. } \lim_{n\rightarrow \infty}X_n(\omega)=X(\omega))=1$
(2) $\lim_{n\rightarrow \infty} P(\omega \in \Omega \text{ s.t. } X_n(\omega)=X(\omega))=1$
So (1) has the limit inside, (2) has the limit outside. 
(2) may look very similar to $X_n\rightarrow_pX$, where $X_n\rightarrow_pX$ means that $\forall \epsilon>0$ $\lim_{n\rightarrow \infty} P(\omega \in \Omega \text{ s.t. } X_n-X\leq \epsilon)=1$
What can we deduce from here?
 A: In fact, $(2)$ is stronger than convergence in probability.  (Assume $(2)$ and take $\epsilon > 0$.  Then $\{X_n = X\} \subseteq \{|X_n - X| < \epsilon\}$.)
Under the given conditions, there are no implications between these statements.  Take $(\Omega, \mathcal{F}, P) = ([0,1], \mathcal{B}([0,1]), \lambda)$. i.e. the Borel sets on $[0,1]$ equipped with the Lebesgue measure.


*

*when both statements are true: simply take $X_n = c$, i.e. be a constant sequence of constant random variable.

*$(1)$ holds but $(2)$ does not: adapt the textbook example for pointwise-convergence.  Take $X_n(\omega) = \omega^n$ and $X \equiv 0$.


*

*$X_n \overset{a.s.}{\to} X$

*$\forall n \in \Bbb{N}, P(X_n = X) = P(\{0\}) = 0$


*$(2)$ holds but $(1)$ does not: adapt the classic "sliding bump" functions for showing that convergence in measure doesn't implies convergence almost everywhere.  Take $Y_{m,n} = 1_{\left[\frac{m-1}{n}, \frac mn \right]}$ for $m \in \{1,\dots,n\}$ and $n \in \Bbb{N}$.  i.e. $Y_{m,n}$ is a "bump $\left[\frac{m-1}{n}, \frac mn \right]$ of width $\frac1n$ sliding from left to right as $m$ runs from $1$ to $n$".  Enumerate $(Y_{m,n})$ as $\{Y_{1,1},Y_{2,1},Y_{2,2}, \dots\}$ and denote this sequence of random variables as $X_k$.


*

*$X_k$ oscillates almost surely as every point on $[0,1]$ is "visited by infinitely many bumps".  For all $\omega \in [0,1]$, choose $n \in \Bbb{N}$.  As the "bumps" $\left[\frac{m-1}{n}, \frac mn \right]$ cover the whole space $[0,1]$, there exists an $m$ so that $Y_{m,n}(\omega) = 1$.  Therefore, $Y_{m,n}(\omega) = 1$ infinitely often.  Idem for $Y_{m,n}(\omega) = 0$, so $(1)$ doesn't hold.

*$P(Y_{m,n} = 0) = \dfrac{n-1}{n} \xrightarrow[n\to\infty]{} 1$, so $(2)$ follows.



Remarks: $(2) \implies (1)$ is "partially true", in the sense that $(2)$ implies convergence in probability implies convergence of a subsequence almost surely.
